Kinematics - motion Flashcards
Deriving v=u + at
Acceleration = gradient of v-t graph
So change in y / change in x = (v-u)/t
a=(v-u)/t
at=v-u
v=u + at
dervive s=1/2 (u+v)t
Displacement = area of v-t graph
Area of trap = (a+b)/2 x h
s= (u+v)/2 x t
s= 1/2(u+v) x t
derive s=ut + 1/2 at^2
sub v=u+at into s=1/2(u+v)t
s=1/2(u+u+at)t
s=1/2(2u+at)t
s=u+1/2at^2
derive s=vt-1/2at^2
rearrange v=u+at to get u= v-at
sun into s=1/2(u+v)t
s=1/2(v-at+v)t
s=1/2(2v -at)t
s=vt -1/2at^2
derive v^2=u^2 +2as
rearrange v=u+at to get t=(v-u)/a
sub into s=1/2(u+v)t
s=1/2(u+v)(v-u/a)
s=1/2a(uv-u^2+v^2-uv)
2as=v^2-u^2
v^2= u^2 + 2as
find velocity on s-t graph
the gradient : displacement /time
kinetmatic equations at constant speed (a=0)
Distance = speed x time
kmph-> ms-1
x1000 dived 60 dived by 60
kinematic equations at constant acceleration
SUVATs in formula booklet
kinematic equations at variable acceleration
S
V
A
down differnetiate
up interstate
express in terms of t
when do you know it’s variable acceleration
expressed in terms of t
speed time graphs at constant speed (a=0)
hortizontal line
d=v x t (area under line)
speed time graphs at constant acceleration
distance and acceleration
use trap rule for distance = area under line
acceleration = gradient of the line
speed time graphs at variable acceleration
shape, distance and accerlatation
curvy line
use intergration for distance = area under the line
acceleration = gradient of line
vector motion at constant speed
calculate position vector
position vector = intial position vector + (speed x time)
vector motion at constant acceleration
find final velocity and displacement
V = u + at
R =intial postion vector + ut + 1/2at^2
R= intial postion vector + vt -1/2at^2
Vector motion at variable acceleration
R
V
A
down differinatiate
up intergrate
acceleration in projectiles
horizontal motion and vertical motion
horizontal: a=0
vertical:a= -g
how to find speed and direction at a certain point on a projectile
both vertical and horizontal use:
Phythagoras for speed
Tan^-1 for direction
vector motion: interpret mathematically “moving in the direction
(2)
(3)”
v= k(2/3)
meaning Parrell to
vecotr motion: interpret “is north east of the orgin”
r=k(1)
(1)
some multiple of
north east position vectors
i=j
north west position vectors
i=-j
south east position vectors
i=-j
South west position vectors
i=j
modelling assumptions: smooth pulley
tension on either side of the pulley is equal
modelling assumptions: light string
tension is equal throughout the string
modelling assumptions: inextensible string
both particles have same acceleration
modelling assumptions: particle
ignore air resistance, ignore rotational effects as no dimensions
modelling assumptions: rod
rigid so it doesn’t bend it has no thickness
